$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}$

106 Views Asked by At

I came across such limit $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}$$ and don't know how to find it. A factor $\frac{1}{n}$ before the sum hints that the limit should be considered as the limit of integral sums but unfortunately I can't select smth like $f(x_k)(x_{k+1} - x_k)$.

Could you please help me to solve it? I would be grateful for any hints. Thanks a lot in advance.

1

There are 1 best solutions below

2
On BEST ANSWER

By Stolz-Cesaro

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}=\lim_{n \to \infty} \frac{n+1}{(n+1)^2 + 1}=0$$